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Chapter 5: Problem 12

The average human lung takes in about 0.6 liter of air with each inhalation, through the mouth and nose, \(A\). This lasts for about 1.5 seconds. Determine the power required to do this if it occurs through the trachea \(B\) having a cross-sectional area of \(125 \mathrm{~mm}^{2}\). Take \(\rho_{a}=1.23 \mathrm{~kg} / \mathrm{m}^{3}\). Hint: Recall that power is force \(F\) times velocity \(V\), where \(F=p A\).

Short Answer

Expert verified
The power required to inhale air through the trachea is given by \(P = \frac{1}{2} * 1.23 kg/m^{3} * (0.4 \times 10^{-2})^{3} * \frac{125}{10^6} m^2\)

Step by step solution

01

Convert units

Because we're using SI units, we first need to convert given units into them. Area of trachea \(A_{trachea}=125 mm^2\), is converted to \(\frac{125}{10^6} m^2\). The air volume inhaled \(V_{inhale}=0.6L\) can be converted to \(0.6 \times 10^{-3} m^{3}\)
02

Calculate the Velocity

Velocity here, \(V\), is the volume inhaled by lungs per unit time. This can be calculated as \(V = \frac{V_{inhale}}{time} = \frac{0.6 \times 10^{-3} m^{3}}{1.5 s} = 0.4 \times 10^{-3} m/s\)
03

Calculate the Pressure

According to Bernoulli's equation, the pressure difference \(p\) can be calculated by \(p= \frac{1}{2}* \rho_{a} * V^2 = \frac{1}{2} *1.23 kg/m^{3} * (0.4 \times 10^{-2})^{2}\)
04

Find the Force

The force exerted in the trachea, \(F\), is calculated by \(F = p * A_{trachea} = \frac{1}{2} *1.23 kg/m^{3} * (0.4 \times 10^{-2})^{2} * \frac{125}{10^6} m^2\)
05

Calculate the Power

Finally, having obtained force \(F\) and velocity \(V\) we find power by using the given hint that power is force times velocity \(P = F*V\). Putting in the values calculated we get, \(P = \frac{1}{2} *1.23 kg/m^{3} * (0.4 \times 10^{-2})^{3} * \frac{125}{10^6} m^2\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bernoulli's Equation
Bernoulli's Equation is fundamental in fluid mechanics and describes the behavior of fluid flow. It states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume for a fluid particle along a streamline is constant. This principle helps us determine pressure differences in various fluid flow scenarios.
In this exercise, we used Bernoulli's equation to calculate the pressure difference required for inhalation. The equation is expressed as:
  • Pressure, \( p = \frac{1}{2} \rho V^2 \)
  • where \( \rho \) is the density of the air, and \( V \) is the velocity of the inhaled air.
By understanding how velocity affects pressure, students can appreciate the dynamics of breathing. Higher velocity results in lower pressure, facilitating air entering the lungs.
Inhalation Mechanics
Inhalation mechanics involves the movement of air into the lungs, driven by pressure differences. This occurs naturally due to the expansion of the thoracic cavity and reduction of pressure in the lungs compared to outside air. For the problem, converting the trachea's cross-sectional area from \(125\ mm^2 \) to \( \frac{125}{10^6} \ m^2\) allows the calculation of airflow velocity.
The velocity of inhaled air is found by dividing the volume of air by the time taken for inhalation:
  • Velocity, \( V = \frac{V_{inhale}}{time} = \frac{0.6 \times 10^{-3} m^{3}}{1.5 s} = 0.4 \times 10^{-3} m/s \)
Inhalation is a fascinating interplay of physics and biology, showing how low pressure draws air efficiently into the respiratory system.
Power Calculation
Calculating the power required for inhalation involves understanding that power is the product of force and velocity. First, we determine the force acting in the trachea using the formula:
  • Force, \( F = p \times A_{trachea} \)
Where \( p \) is the pressure calculated from Bernoulli's equation.
Finally, power is calculated by multiplying this force by the velocity of air:
  • Power, \( P = F \times V \)
  • This gives the rate at which energy is used to move air during inhalation.
The calculation of power demonstrates how much effort is required by the respiratory system to perform the inhalation task. Such an understanding can be helpful not only in physics, but also in physiological studies, highlighting the efficiency of human breathing.

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Most popular questions from this chapter

An ideal fluid having a density \(\rho\) flows with a velocity \(V\) through the horizontal pipe bend. Plot the pressure variation within the fluid as a function of the radius \(r,\) where \(r_{i} \leq r \leq r_{o}\) and \(r_{o}=2 r_{i} .\) For the calculation assume the velocity is constant over the cross section.

Water is delivered from one reservoir to another at a height of \(18 \mathrm{~m}\). If the friction head loss in the connection pipe is \(3 \mathrm{~m}\) per kilometer of the pipe. The diameter and the length of the connection pipe are \(180 \mathrm{~mm}\) and \(3 \mathrm{~km}\) long, respectively. Determine the required power output of a pump to maintain the flow of \(0.6 \mathrm{~m}^{3} / \mathrm{s}\). The ends of the pipe are submerged in the reservoir.

The power input of the pump is \(10 \mathrm{~kW}\) and the friction head loss between \(A\) and \(B\) is \(1.25 \mathrm{~m}\). If the pump has an efficiency of \(e=0.8,\) and the increase in pressure from \(A\) to \(B\) is \(100 \mathrm{kPa}\), determine the volumetric flow of water through the pump.

The pump discharges water at \(B\) at \(0.05 \mathrm{~m}^{3} / \mathrm{s}\). If the friction head loss between the intake at \(A\) and the outlet at \(B\) is \(0.9 \mathrm{~m},\) and the power input to the pump is \(8 \mathrm{~kW}\) determine the difference in pressure between \(A\) and \(B\). The efficiency of the pump is \(e=0.7\).

Oil flows through the constant-diameter pipe such that at \(A\) the pressure is \(50 \mathrm{kPa}\), and the velocity is \(2 \mathrm{~m} / \mathrm{s}\). Plot the pressure head and the gravitational head for \(A B\) using a datum at \(B\). Take \(\rho_{o}=900 \mathrm{~kg} / \mathrm{m}^{3}\).
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